First slide

Methods of integration

Question

If $\int \frac{d x}{1 + e^{x}} = x + Klog (1 + e^{x}) + C$ then $K$ is equal to

Easy

A

1
B

-1
C

2
D

-2

Solution

Substituting $1 + e^{x} = t$ we have
$\begin{array}{l} \int \frac{d x}{1 + e^{x}} = \int \frac{d t}{t (t - 1)} = \int \frac{1}{t - 1} d t - \int \frac{d t}{t} \\ = \log [t - 1] - \log t + C \\ = \log e^{x} - \log (1 + e^{x}) + C \\ = x - \log (1 + e^{x}) + C \end{array}$

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